Max/Min question help (1 Viewer)

[5647382910]

The cost of running a car at an average speed of v km/h is given by c = 150 + v^2/80 cents per hour. Find the average speed, to the nearest km/, at which the cost of a 500km trip is a minimum.

Answer: 110km/h

I just cant seem to do this q for some weird reason.

 

[midifile]

The cost of running a car at an average speed of v km/h is given by c = 150 + v^2/80 cents per hour. Find the average speed, to the nearest km/, at which the cost of a 500km trip is a minimum.

Answer: 110km/h

I just cant seem to do this q for some weird reason.

Cost per hour = 150 + v^2/80
Therefore total cost (T) = (150 + v^2/80) x (500/v)
T = 75000v^-1 + 6.25v
therefore dT/dv = -75000v^-2 + 6.25

Minimum cost occurs when dT/dv = 0
therefore -75000v^-2 + 6.25 = 0
v^2 = 12000
therefore v = +-109.54 = +-110 km/hr
But v>0, so b = 110 km/hr

 

[Aerath]

Do we have to prove that it is a max/min?

 

[midifile]

Yeah, you normally do I just couldnt be bothered typing it out